Question

Log2(3x-2)<log2(2x-3)​

Answer 1

ОДЗ 3х - 6 > 0 => х > 2 2х - 3 > 0 => х > 1,5

3х - 6 = 2х - 3

2х - 2х = -3 + 6

х = 3

Answer 2

To solve the inequality log2(3x - 2) < log2(2x - 3), you can use the properties of logarithms.

First, remember that for any positive number "a," loga(b) < loga(c) if and only if b < c. In this case, a is 2.

So, you can rewrite the inequality as:

3x - 2 < 2x - 3

Now, solve for x:

3x - 2 - 2x + 3 < 0

x + 1 < 0

To isolate x, subtract 1 from both sides of the inequality:

x < -1

So, the solution to the inequality log2(3x - 2) < log2(2x - 3) is:

x < -1